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Ex 10.4, 9 - Chapter 10 Class 12 Vector Algebra
Last updated at April 16, 2024 by Teachoo
Chapter 10 Class 12 Vector Algebra
Concept wise
- Scalar or vector
- Graphical displacement
- Type of vector
- Multiplication of a vector by a scalar
- Equal vectors
- Unit vector
- Direction cosines and ratios
- Addition(resultant) of vectors
- Joining two points
- Section formula
- Right Angled triangle
- Collinearity Of two vectors
- Collinearity of 3 points or 3 position vectors
- Scalar product - Defination
- Scalar product - Projection
- Scalar product - Solving
- Vector product - Defination
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Vector product - Area
- Vector product - Solving
Transcript
Ex 10.4, 9 Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5). A (1, 1, 2) , B (2, 3, 5) , C (1, 5, 5) Area of triangle ABC is 1/2 |(𝐴𝐵) ⃗ × (𝐴𝐶) ⃗ | A (1, 1, 2) B (2, 3, 5) (𝐴𝐵) ⃗ = (2 − 1) 𝑖 ̂ + (3 − 1) 𝑗 ̂ + (5 − 2) 𝑘 ̂ = 1𝑖 ̂ + 2𝑗 ̂ + 3𝑘 ̂ A (1, 1, 2) C (1, 5, 5) (𝐴𝐶) ⃗ = (1 − 1) 𝑖 ̂ + (5 − 1) 𝑗 ̂ + (5 − 2) 𝑘 ̂ = 0𝑖 ̂ + 4𝑗 ̂ + 3𝑘 ̂ (𝑨𝑩) ⃗ × (𝑨𝑪) ⃗ = |■8(𝑖 ̂&𝑗 ̂&𝑘 ̂@1&2&3@0&4&3)| = 𝑖 ̂ (2×3−4×3 )−𝑗 ̂ (1×3−0×3 )+𝑘 ̂ (1×4−0×2 ) = 𝑖 ̂ (6−12)−𝑗 ̂(3−0) + 𝑘 ̂ (4−0) = −6𝒊 ̂ – 3𝒋 ̂ + 4𝒌 ̂ Magnitude of (𝐴𝐵) ⃗ × (𝐴𝐶) ⃗ = √((−6)2+(−3)2+42) |(𝐴𝐵) ⃗" × " (𝐴𝐶) ⃗ | = √(36+9+16) = √61 Area of triangle ABC = 1/2 |(𝐴𝐵) ⃗" × " (𝐴𝐶) ⃗ | = 1/2 × √61 = √𝟔𝟏/𝟐
